3 votes 3 votes Compute approximate optimal window size when packet size is 53 bytes, the RTT is 60 micro seconds and bottleneck bandwidth is 155 Mbps? Computer Networks computer-networks + – focus _GATE asked Dec 16, 2016 focus _GATE 3.0k views answer comment Share Follow See all 9 Comments See all 9 9 Comments reply Show 6 previous comments stanchion commented Oct 26, 2017 reply Follow Share Yes, bibhas kumar is right. AS per Sliding window is concern, window size is calc. by: Window size= 1+2a = floor(1+ 21.93) = floor(22.93) = 22 Correct me if i am wrong. 0 votes 0 votes Groot commented Jan 2, 2018 reply Follow Share refer https://gateoverflow.in/1820/gate2006-44 0 votes 0 votes Venkat Sai commented Jan 21, 2018 reply Follow Share @stanchion its not mentioned that its sliding window so dont assume what digvijay singh said is right 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes BD product= RTT*Bandwidth =155∗106∗60∗10−6=155*60 packet size 53*8 so optimal window size=(155*60)/(53*8)=21.93 =22 Aboveallplayer answered Dec 26, 2016 Aboveallplayer comment Share Follow See all 2 Comments See all 2 2 Comments reply Ravit Anand commented Oct 9, 2017 reply Follow Share Why you are taking 22? 0 votes 0 votes mohitbawankar commented Dec 11, 2017 reply Follow Share Window size= floor ( BW*RTT / packet size) =floor (155∗10^6∗60∗10^−6/53*8) = floor(21.93)=21 0 votes 0 votes Please log in or register to add a comment.
